Solution to Introduction to Modern Analysis by Edward D. Gaughan, Fifth Edition.
ID: 2985577 • Letter: S
Question
Solution to Introduction to Modern Analysis by Edward D. Gaughan, Fifth Edition. Chapter 0, Question 41.
0<a^2<b^2
Proof:
given 0<a<b we see that
a<b implies that b-a>0
also since a and b >0, this implies that
a+b>0 and b+a>0
from these two facts we observe the following relation:
(b+a)(b-a)>0
b^2+ab-ab+a^2
b^2+a^2
this gives proof conditional that
a>0
(a)(a)>0(a)
a^2>0
whence: 0<a^2<b^2 Q.E.D.
___________________________________________________________________
0<sqrt(a)<sqrt(b)
sqrt(a)<sqrt(b) implies that sqrt(b) - sqrt(a) > 0
also since sqrt(a) and sqrt(b) > 0, this implies that
sqrt(a) + sqrt(b) > 0 and sqrt(b) + sqrt(a) > 0
from these two facts we observe the following relation:
(sqrt(b) + sqrt(a)) (sqrt(b) - sqrt(a)) > 0
b + sqrt(ab) - sqrt(ab) - a > 0
b-a > 0
b > a
this gives converse proof conditional that
sqrt(a) > 0
sqrt(a)sqrt(a) > 0
a > 0
whence: 0<sqrt(a)<sqrt(b) Q.E.D.
Explanation / Answer
if nobody answers plz rate me plzzzzzzzzzzzz
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.